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Unformatted text preview: MasteringPhysics: Assignment Print View http://sessiortmasteringphysics.com/myct/assignmentPrint?assigrimen.. Assignment Display Mode: l: PHYSICSBCFOB ctric catechism it fawn??? m2 tsécmztay, Qctomr 23%8 \ﬁew Grading Details The Change Inside a Conductor Description: A spherical conductor has a cavity containing a ﬁxed charge. Comeptual questions about how charge is arranged on the surfaces of a conductor to cancel the ﬁeld due to
i arrangement would change if an additional external charge were brought near the conductor. A spherical cavity is hollowed out of the interior of a neutral condueting sphere. At the center of the cavity is a point charge, of positive charge 1;. m.r.rrwt.ma«an«mm\mnwdm.m ,3 Part A
, What is the total surface charge rim, on the interior surface of the conductor (i.e., on the wall of the cavity)? Hint A.l Gauss's law and properties of conductors The net electric ﬁeld in the interior of the conducting material must be zero. (The electric ﬁeld in the cavity, however, need not be zero.) Knowing this, you can use Gauss's law to ﬁnd the net charge on the interior surface of the
‘ cavity. Use the following Gaussian surface: an imaginary sphere, centered at the cavity, that has an inﬁnitesimally larger radius than that of the cavity, so that it encompasses the inner surface of the cavity. This Gaussian surface
' lies within the conductor, so the ﬁeld onthe Gaussian surface must be zero. Thus, by Gauss's law, the net charge inside the Gaussian surface must be zero as well. But you know that there is a point charge :; within the Gaussian urface. If the net charge within the Gaussian surface must be zero, how much charge must be present on the surfa of the (has: “"1 13 Part B
I What is the total surface charge am on the exterior surface of the conductor? Properties of the conductor i = In the problem introduction you are told that the conducting sphere is neutral. Furthermore, recall that the free charges within a conductor always accumulate on the conductor's surface (or surfaces, in this case). You found the net
harge on the conductor’s interior surface in Part A. If the conductor is to have zero net charge (as it must, since it is neutral), how much charge must be present on its exterior surface? ‘ % ANSWER: : (gm = f; Part C
What is the magnitude Em of the electric ﬁeld inside the cavity as a ﬁmction of the distance a” from the point charge? Let it, as usual, denote ‘n‘ u Hint C.1 How to approach the problem : The net electric ﬁeld inside the conductor has three contributions: . 1, fromthe charge q; 2. ﬁ'omthe charge on the cavity's walls 5;”;
E 3. from the charge on the outer stn‘face of the spherical conductor gm. However, the net electric ﬁeld inside the conductor must be zero. How must QM and am be distributed for this to happen? Here's a clue: the ﬁrst two contributions above cancel each other out, outside the cavity. Then the electric ﬁeld produced by W inside the spherical conductor must separately be zero also. How must {am be distributed for this
i o happen? ‘ E Aﬁer you have ﬁgured out how (gm and W are distributed, it will be easy to ﬁnd the ﬁeld in the cavity, either by adding ﬁeld contributions ﬁom all charges, or using Gauss's Law. V i 1 Part 02 Charge distributions and ﬁnding the electric field : i and gm are both uniformly distributed. Unfortunately there is no easy way to detemiine this, that is why a clue was given in the last hint. You might hit upon it by assuming the simplest possible distribution (i.e., uniform) or
’ by trial and error, and check that it works (gives no net electric ﬁeld inside the conductor). If gm is distributed uniformly over the surface of the conducting sphere, it will not produce a net electric ﬁeld inside the sphere. What are the characteristics of the ﬁeld em, produces inside the cavity? ANSWER: H I . zen, V
the same as the ﬁeld produced by a point charge {3’ located at the center of the sphere loflO 10/31/200811:20AM MasteringPhysics: Assignment Print View http://session.masteringphysics.conﬂmycﬂassigmnentPrinthssignmen... the same as the ﬁeld produced by a point charge located at the position of the charge in the cavity ANSWER: 0 Part D
What is the electric ﬁeld Em outside the conductor? : Hint DJ How to approach the problem
The ret electric ﬁeld inside the conductor has three contributions: L from the charge :3;
2. from the charge on the cavity‘s walls q“;
3. fromthe charge on the outer surface of the spherical conductor arm. : However, the net electric ﬁeld inside the conductor must be zero. How must (rm, and M be distributed for this to happen? : Here's a helpful clue: the ﬁrst two contributions above cancel each other out, outside the cavity. Then the electric ﬁeld produced by gm inside the spherical conductor must be separately be zero also. How must gm be
I distributed for this to happen‘7 What sort of ﬁeld would such a distribution produce outside the conductor? t 'Hint 0.2 The distribution of QM . If is distributed uniformly over the surface of the corﬁucting sphere, it will not produce a net electric ﬁeld inside the sphere. What are the characteristics of the ﬁeld it produces outside the sphere? ANSWER: ' we
. the same as the ﬁeld produced by a point charge {3! located at the center of the sphere
the same as the ﬁeld produced by a point charge located at the position of the charge in the cavity 4 Now a second (352, is broughtnear the outside of the conductor. Which of the following quantities would change? PartE The total surface charge on the wall of the cavity, 1;“: Canceling the ﬁeld due to the charge (,3 H would change
.wouid not change f The total surface charge on the exterior of the conductor, gm: Canceling the ﬁeld due to the charge :33 The net electric ﬁeld inside a conductor is always zero. The charges on the outer surface of the conductor will rearrange thetmelves to shield the external ﬁeld completely. Does this require the net charge on the outer surface to
> change? i
i
E ANSWER: would change
. would not change Part G
~ The electric ﬁeld withinthe cavity, ﬂy“: ANSWER: would change
. would not change The electric ﬁeld outside the conductor, 15;“: ANSWER: : . would Change
would not change Description: A coaxial cable consists of an inner wire and a concentric cylindrical outer conductor shown in the ﬁgure . (a) If the conductors carry equal but opposite charges, show that there is no net charge on the outside of the
outer conductors coaxial cable consists of an inner wire and a concentric cylindrical outer conductor shown in the ﬁgure . 2 oflO 10/31/200811:20 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.cont/myct/assigmnentPrinﬂassignmen... l
i é.
é If the conductors carry equal but opposite charges, show that there is no net charge onthe outside of the outer conductor.
ANSWER: AmwerKey:
'Assume line symmetry, and apply Gauss's law to the outer cylindrical conducting surface, 2*pi*r*L*E_.suf = q_enclosed/epsilon’0. Since the conductors in the cable carry opposite charges of equal magnitude, there is zero charge enclosed, so the field and the
Icharge density there (sigma = epsilon~O*E_suf) vanish. ric Potential g Electric Potential Energy versus Elect Description: Introduces concept of electric potential energy and its relationship to the electrostatic force by analogy with the gravitational potential/force. Compares electric potential energy with electric potential. Learning Goal: To understand the relationship and differences between electric potential and electric potential energy. n this problem we will learn about the relationships between electric force £3, electric ﬁeld 1?, potential energy if, and electric potential 1". To understand these concepts, we will ﬁrst study a system with which you are already miliar: the uniform gravitational ﬁeld. Gravitational Force and Potential Energy i i First we review the force and potential energy of an object of mass g in a uniform gravitational ﬁeld that points downward (in the ~— 2 direction), like the gravitational ﬁeld near the earth's surface. Part A Find the force I} {,3; on an object of mass 1;; inthe uniform gravitational ﬁeld when it is at height a m 6. Express 13?{;§interms of m, 2, g, and g. ANSWER: f pm: _m§,5 Because we are m a uniform ﬁeld, the force does not depend on the object's location Therefore, the variable a does not appear in the correct answer. :55 Part B Now ﬁnd the gravitational potential energy if {a} of the object when it is at an arbitrary height 3. Take zero potential to be at position 3, Express {3 ;:§ in term of m, z, and g. Note that became potential energy is a scalar, and not a vector, there will be no unit vectorin the answer. ANSWER: #53 {ti = mg: Part C In what direction does the object accelerate when released with initial velocity upward? ANSWER: upward
. downward upward or downward depending on its mss
dowuward only if the ratio of g to initial velocity is large enough Electric Force and Potential Energy Now consider the analogous case of a particle with charge 3 placed in a uniform electric ﬁeld of strength R7, pointing downward (in the m E direction) Part D
Find if; z}, the electric force on the charged particle at height a. nt D.1 Relationship between force and electric ﬁeld V The force on a particle of charge t; in an electric ﬁeld Express 3.; in term of r}, 13}, a, and 3 of10 10/31/200811:20 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigntnentPrint?assignmen... ANSWER: : xix}: Wm is? Part E
Now ﬁnd the potential energy {I {a} of this charged particle when it is at height 3. Take zero potential to be at position 3 as! it. Express {23% (a scalar quantity) in terms of 1;, 1;, and ;. ANSWER:  {3:3} = (Egg: Part F
; In what direction does the charged particle accelerate when released with upward initial velocity? ANSWER: upward downward _ ‘ upward or downward depending on its charge 3 downward only if the ratio of of}; to initial velocity is large enough S E Electric Fleld and Electric Potential The Elem“: 13°tential V is deﬁned by the rehubris“? U :13", Where is the electric potential energy of a particle with charge (3. i Part G
3 V Find the electric potential V of the uniform electric ﬁeld 3%. Note that this ﬁeld is not pointing in the same direction as the ﬁeld in the previous section of this problem Take zero potential to be at position 7; W: V Express 3" interns of :1, E3, and e. ANSWER: : 1; = who)“: Part H
a The SI unit for electric potential is the volt. The volt is a derived unit, which means that it can be written in terms of other SI units. What are the dimensions of the volt in terms of the ﬁmdamental SI units? Express your answerin terms of the standard abbreviations for the fundamental SI units: in (meters), Egg (kilograms), 2; (seconds), and (3 (coulombs) ANSWER: volts = Part I ‘ E i The electric ﬁeld can be derived fromthe electric potential, just as the electrostatic force can be determined from the electric potential energy. The relationship between electric ﬁeld and electric potential is m w ‘6‘? , where is the gradient operator: a y 81% (WV, 695“
Vi» m at «it get. ‘ a I The partial derivative means the derivative of V with respect to x, holding all other variables constant.
3 (he:
: I 1' I; If y p ) »
Consider again the electric potential V :2; ~12”; corresponding to the ﬁeld If: 13226. This potential depends ontl‘ie z coordinate only, so rm :w t} and » t 5 t’ 3 Find an expression for the electric ﬁeld in terms of the derivative of V.
Express your nnsweras a vectorin tenm of the unit vectors ﬁr, 3}, and/or ,Q. Use iii/fa}; for the derivative of V with respect to a. ANSWER: Part J
\, A positive test charge will accelerate toward regions of electric potential and electric potential energy. Direction of the electric ﬁeld what direction do electric ﬁelds point? ANSWER: ‘ from regions of higier electric potential to lower electric potential
from regions of lower electric potential to higher electric potential Formula for the force on a charge in an electric ﬁeld 40f10 10/31/2008 11:20AM MasteringPhysics: Assignment Print View http://session.masteringphysics.corn/myct/assigmnentPrinﬂassignmen... EThe force [5: on a charge (31 in anelectric ﬁeld is given by Fm This is similar to the equation F mg for the force on a mass in a uniform gravitational ﬁeld. V Hint J.3 Formula for electric potential energy
‘ The electric potential energy if of a charge (3 at electric potential V is given by i; This is similar to the equation I‘ ’ m high, for the gravitational potential energy of a particle with mass to. Choose the appropriate answer combination to ﬁll in the blanks correctly. ANSWER: , higher; higher
I higher; lower lower; higher
. lower; lower Part K
A negative test charge will accelerate toward regions of electric potential and electric potential energy. Part K.1 Direction of the electric field In what direction do electric ﬁelds point? ANSWER: Q from regions of higher electric potential to lower electric potential
from regions of lower electric potential to higher electric potential Hint K2 Formula for the force on a charge in an electric ﬁeld The force if on a charge :3 in an electric ﬁeld E? is given by This is similar to the equation F mg for the force on a mass in a uniform gravitational ﬁeld. Hint K3 Formula for electric potential energy
The electric potential energy I} of a charge :1 at electric potential V is given by his is similar to the equation {2 Choose the appropriate answer combination to ﬁll in the blanks correctly. ANSWER: > higher; hing
6 higher; lower lower; higher lower; lower E A charge in an electric ﬁeld will experienc
'rection of decreasing electric potential energy is the direction of increasing electric potential. Potential Energy of a Battery Description: The electric potential and potential energy associated with a battery. Learning Goal: To understand electrical potential, electrical potential energy, and the relationship between them
ms. Electrical potential energy {I}; is the potential energy that a charge I; has due to its position relative to other arge a particle of net charge r; would have at that position. In other words. if a charge q has an electric potential Electric potential and electric potential energy are related but diﬁ‘erent concepts. Be careful not to confuse the ter
i charges. The electric potential V at a speciﬁc position is a measure of the amount of potential energy per unit ch E
E
t ‘2 energy {ﬁt , the electric potential V at the location of x; is “‘2‘. Recall that the gravitational potential energl (5:12 a: orgy) of an object of mass m depends on where you deﬁne y {3. The diﬂerence at?“ in gravitational potential energy between two points is the physically relevant quantity. ‘> Similarly. for electric potential energy, the important quantity is the change in electric potential energy: ALI}; ygrkir‘. This is why we oﬂen just measure the potential difference at”. When we say that the potential of a car battery is 12 \3, we mean that the potential difference between the positive and negative terminals of the battery is 12 V. i
i i
g i Consider dropping a ball from rest. This ball moves from a state of high gravitational potential energy to one of low gravitational potential energy as it falls to the ground. Similarly, charges move ﬁ'om a state of high electric potential energy to one of low electric potential energy. Part A 5 0f10 10/31/200811:20 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigrimentPrint?assignmen.. I Mustang Sally just ﬁnished restoring her 1965 Ford Mustang car, To save money, she did not get a new battery. When she tries to start the car, she discovers that the battery is dead and she needs ajump start. While unhooking the
jumper cables, the positive and negative cables almost touch and a sparkjumps between the ends of the cables. This spark is caused by the movement of electrons through the air between the battery terminals. In what direction are \ the electrons traveling? Hint A.1 Another way to think about the movement of charge You can think of the movement of charges in terms of Coulomb's force. A positive (high) potential is created by positive charges and a low (negative) potential is created by negative charges To understand which way electrons
» will ﬂow across a potential difference, think about the forces on an electron. An electron will be repelled by a negative charge and attracted to a positive charge. The negative terminal of a battery can be viewed as having a negative charge. The posmve terminal is at a higher potential than thewnegative terminal. Unless provided with energy, pos ve charges will ﬂow from a high to a low potential, and negatively charged electrons will ﬂow from a low to a high
potential. The table below summarizes this movement. Direction of motion high to low potential low to high potential $5“Since potential difference is the energy per unit charge, it is measm'ed in units of energy divided by charge. Speciﬁcally, potential difference is generally measured in volts (whose symbol is Y). One volt is equal to one joule
E per coulomb: E V w 1 351 (ff. There is a 12 V potential difference between the positive and negative ends of the jumper cables, which are a short distance apart. An electron at the negative end ready to jinnp to the positive end has a certain amount of potential
energy. On what quantiﬁes does this electrical potential energy depend? Him 3.1 The expression for electric potential energy The electric potential energy dilference is given by the potential difference times the charge :3. ’ ANSWER: ; the distance between the ends ofthe cables
' the potential difference between the ends of the cables
the charge on the electron
the distance and the potential difference
the distance and the charge
. the potential diiference and the charge
the potential diEerence, charge, and distance E ‘53 Part C
Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom There is now a negative ion with a charge 2!: at this terminal. What are the electric potential and electric potential energy of the negative ion relative to the electron? i
z ANSWER: The electric potential and the electric potential energy are both twice as much.
The electric potential is twice as much and the electric potential energy is the same.
. The electric potential is the same and the electric potential energy is twice as much
The electric potential and the electric potential energy are both the same.
The electric potential is the same and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron
The electric potential is twice as much and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron. é ‘yg Part D
What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable? In other words, assume that the electric potential of the positive terminal is O V and that of the negative 1 terminal is :2 Vi Recall that a a» too it. to“ (,3. Enter your answer numerically in joules. ANSWER: ‘ = ‘E‘m‘igmth J W Part E
At the negative terminal of the battery the electron has electric potential energy. What happens to this energy as the electron jumps from the negative to the positive tenniml? ANSWER: ; It disappears.
Q It is converted to kinetic energy. It heats the battery.
It increases the potential of the battery. [51 as gravitational potential energy is converted to kinetic energy when something falls, electrical potential energy is converted to kinetic energy when a charge goes froma high potential energy state to a low potential
nergy state. 3? Part F
If you wanted to move an electron from the positive to the negative terminal of the battery, how much work W would you need to do on the electron? 6 oflO 10/31/2008 11:20 AM MasteringPhysics: Assignment Print View http://sessi0n.masteringphysics.com/myct/assignmentPrint?assignmen... Hint F.1 Formula for work
The work done on a charge (; is equal to the product » 9:31" Enter your answer numerically in joules. ANSWER: ‘ “a = L_§}g,§emxx } g Because moving a negative charge fromthe positive to the negative terminal of the battery would increase its electric potential energy, it would take positive work to move the charge. This is simliar to liﬁing a ball upward.
5 You do positive work on the ball to increase its gravitational potential energy. E
§ § E W Electric Fields and Equipotential Surfaces : Description: Find the work done to move a unit charge ﬁ'om and to given points on a diagram showing equipotential surfaces, and compare the magnitude of the electric ﬁeld at these points. E The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in lV increments. E 3 i I V % mg V
—3 V Part A
What is the work it)“; done by the electric force to move a 1—(3 charge from A to B? Part A.l Find the potential difference between A and B l H
What is the potential difference VA m if}; between point A and point B? Hint A.l.a Equlpotential surfaces Recall that an equipotential surface is a surface on which the electric potential is the same at every Express your answer in volts. ANSWER: VA me}; = g} t; 3Hint A.2 Potential difference and work Recall that the potential difference (in volts) between a point a and a point it equals the work (in joules) done by the electric force to move a 14:} charge from a to 5. Express your answer in joules. ANSWER: ‘ t4th» 2 n (1 Part B
What is the work “’51) done by the electric force to move a 10 charge from A to D? i
§
§ Part 3.1 Find the potential difference between A and D
What is the potential difference V», m it}: between point A and point D? xpress your answer in volts. ‘ charge from a to or Express your answer in joules. ANSWER: mm :1 3 Part C
‘e The magnitude ofthe electric ﬁeld at point C is Electric ﬁeld and eqtupotentral surfaces inee the diagram shows equal potential differences between adjacent surfaces, equal amounts of work are done to trove a particular charge from one surface to the next adjacent one. It follows then that if the equipotentials are
loser together, the electric force does the same amount of work in a smaller displacement than if the equipotentials were farther apart. Therefore, the electric force, as well as the corresponding electric ﬁeld, lets a larger 7 of10 10/31/200811:20 AM MasteringPhysics: Assignment Print View http://sessiortmasteringphysics.com/rnyct/assignmentPrint?assignmen... ANSWER: . greater than the magnitude of the electric ﬁeld at point B.
' less than the magnitude of the electric ﬁeld at point B.
equal to the magnitude of the electric ﬁeld at point B.
unknown because the value of the electric potential at point C is unknown. Description: Compute the electric potential on the z axis due to a uniformly charged ring in the xy plane. Then compute the electric ﬁeld on that axis. Potential of a Charged Ring A ring with radius It, and a uniformly distributed total charge Q lies in the xy plane, centered at the origin z Part A . What is the potential V {:3 due to the ring onthe z axis as a function of :2? : ; Hint A.l How to approach the problem : The formula for the electric potential produced by a static charge distribution involves the amount of charge and the distance ﬁ'omthe charge to the position where the potential is measured. All points on the ring are equidistant
ﬁ'om a given point on the z axis. This enables you to calculate the electric potential simply, without doing an integral. Hint A.2 The potential due to a point charge . If you incorporate the symmetry of the problem. you will need only to know the formula for potential of a point charge: Vita} m M, where t/ {at} is the potential at distance ;' from the point charge, I; is the magnitude of the
mar fcharge, and re is the permittivity of free space.
Express your answer in terms of Q, 3, R, and is} or & 1': x am ANSWER: i Part B
What is the magnitude ofthe electric ﬁeld onthe z axis as a ﬁntction of ,2, for 2 > {3? ran 13.11 iiiiii H V Determine the direction oftheﬂfield By symmetry, the electric ﬁeld has only one Cartesian component. In what direction does the electric ﬁeld point? ANSWER: 2w ' xv: Hint 13.2 The relationship between electric ﬁeld and potential You can obtain the electric ﬁeld from a potential by the following expression: a; where is the electric ﬁeld vector, V is the electric potential, is the gradient operator, and means the partial derivative of g“ with respect to g.
r ANSWER:  and then starts to decrease. 8 of10 10/31/2008 11:20 ‘AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen... ‘ Why does the electric ﬁeld exhibit such a behavior?
Though the contribution to the electric ﬁeld ﬁom each point on the ring strictly decreases as a function of z, the vector cancellation from points on opposite sides of the ring becomes very strong for sn'all ;. 134‘ {i} i} on account of these vector cancellations On the other hand gig w x} we 0, even though all the individual gigs point in (almost) the same direction there, because the contribution to the electric ﬁeld, per unit length of the An Electron in a Diode Description: Calculate the ﬁnal speed of an electron (initially at rest) that travels from the cathode to the anode in a cylindrical diode, given the potential diﬁercnce between the electrodes 2 Before the advent of solidstate electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. g
1 One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode. with the anode at the higher potential. art A
t Suppose a diode consists of a cylindrical cathode with a radius ot‘6.2()0><10_2 (sin, mounted coaxially within a cylindrical anode with a radius of 0.5580 (an. The potential difference between the anode and cathode is 265 V. An
i electron leaves the surface of the cathode with zero initial speed (whim; 0). Find its speed rm; when it strikes the anode. % Hint A.1 How to approach the problem
Try to draw a simple diagram of the diode, with the path of the electron going trom the central cathode to the outer anode. Since the diode is a cylinder, the symmetry implies that only radial motion needs to be considered for the
electron, since any motion of the electron around the center will not change its potential energy. Note that only a potential difference is given between the plates, so you will need to be careﬁil about your deﬁnitions. Use the equation for the conservation of energy to ﬁnd the ﬁnal speed of the electron. 1‘3 Pan A,2 Calculate the change in potential energy of the electh
Calculate the change in the potential energy 33!} of the electron as it moves from the inner cathode to the outer anode. Hint Ala
Recall that the potential energy if“ of a charge 1; in an electric ﬁeld is related to the potential i3 of that ﬁeld, evaluated at the position of the charge, by the equation 5 7Q m gig. In this problem, the potential is not known at ’ either the anode or cathode. However, the potential difference AV me w ﬁﬂb‘m is given, so use the relation above to ﬁnd the change in the potential energy :31? of the electron Potential energy and potential Express your answer numerically in joules. ANSWER: AI} = 3 76 Pan A3 Calculate the initial kinetic energy of the electron Calculate the initial kinetic energy KW of the electron at the cathode. Initial velocity of the electron
; Recall that the electron is emitted from the surface f the tired Express your answer in joules. ANSWER: = g, 3 A new W Hint AA Putting it all together i
i The equation of conservation of energy, KNEW «a ﬁrm“ as Kmmg a £58,,” can be rewritten as Km m KWM {Sam m {few m Km“ — Aéi. Since K :2 rm? you now have enough inforrmtionto ﬁnd the ﬁnal peed of the electron. 3 Express your answer numerically in meters per second. ANSWER: Note that the size of the diode makes no difference, as long as the potential difference between the two electrodes is a known constant. Also, note that the potential at the surface of the anode and cathode are not known
separately, but the potential difference is enough for these calculations. In general, the potential at a particular point is not physically important Only potential differences are important, just as only the change in potential
3. \\ energl is important to mechanics problem. WW E . Eroblem 2236 9 oflO ’ 10/31/2008 11:20 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myctmssignmentPrint?assignmen... Description: The ﬁgure shows a uniform electric ﬁeld of magnitude E. Find expressions for the potential diﬁerences (a) Delta V_AB and (b) Delta V_BC . (c) Use your results to determine Delta V_AC. (a) (b) (c) The ﬁgure shows a uniform electric ﬁeld of magnitude 2‘52 Find expressions for the potential differences (a) Aiﬁg and (b) Ai’m . (c) Use your results to determine {Mimi i Part A Express your answer in terns of the variables {3’ and if. Part B Express your answer in terms of the variables :35 and it. ( Express your answerin terms of the variables I? and :1. ANSWER:
QVA‘: : (1 “M Prob]; 22.47 Description: (a) A solid sphere of radius R carries a net charge Q distributed uniformly throughout its volume. Find the potential difference from the sphere's surface to its center. Part A A solid sphere of radius It carries a net charge Q distributed unifonnly throughout its volume. Find the potential difference from the sphere's surface to its center. Express your amwer in tenm of the variables 12, Q and Coulomb constant iv. ANSWER: . , ,
mg} — s» {it} = “Problem22.73 ii V H W Description: A uranium nucleus (mass 238 u, charge 92e) decays, emitting an alpha particle (mass 4 u, charge 2e) and leaving a thorium nuclem (mass 234 u, charge 90c). At the instant the alpha particle leaves the nucleus, the
centers of the two are ## ﬁn apart... g
i " A uranium nucleus (mass 238 it, charge 92 :2) decays, emitting an alpha particle (mass 4 :5, charge 25,) and leaving a thorium nucleus (mass 234 n, elmge 90¢). At the instant the alpha particle leaves the nucleus, the centers of the
two are 4.0 $35; apart and essentially at rest. Part A Find their speeds when they're a great distance apart. Treat each particle as a spherical charge distribution. Express your answers using two signiﬁcant ﬁgures. ANSWER‘ WM: l'i»2§§i§"§‘}{}7 masters refs E Mommwmms lllllllllllll mmmm \\\\\\\\\\\ WWW» : Summary v items complete score) 3 10 of10 10/31/200811:20AM ...
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This note was uploaded on 03/08/2011 for the course PHYS 6c taught by Professor Smith,d during the Spring '08 term at UCSC.
 Spring '08
 Smith,D
 Physics

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